Measuring the size and shape of macromolecules. Hydrodynamics: study of the objects in water How do the move? Translation Rotation

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1 Measuring the size and shape of macromolecules Hydrodynamics: study of the objects in water How do the move? Translation Rotation 1) Movement with no external forcefree diffusion 2) Movement under the influence of an external forcee.g. centrifuge electrophoresis

2 Hydrodynamics techniques measure the frictional resistance of the moving macromolecule in solution: 1. Intrinsic viscosity the influence of the object on the solution properties 2. Free translational diffusion 3. Centrifugation/sedimentation 4. Electrophoresis 5. Gel filtration chromatography 6. Free rotational diffusion/fluorescence anisotropy

3 Hydrodynamics Measurements of the motion of macromolecules in solution form the basis of most methods used to determine molecular size, density, shape, molecular weight Equations based on the behavior of macroscopic objects in water can be used successfully to analyze molecular behavior in solution Techniques sphere ellipsoid disk rod (prolate, oblate) etc. Free Motion Mass Transport Viscosity translational diffusion sedimentation rotational diffusion electrophoresis gel filtration

4 The hydrodynamic particle hydrodynamically bound water shear plane: frictional resistance to motion depends on size shape

5 Hydrodynamic Parameters 1. Specific volume: (Inverse of density) volume = gram V i 1 = solvent 2 = solute Volume = V 1 g 1 + V 2 g 2 vol x grams gram Proteins: V 2 ~ ml/g Na + DNA: V 2 ~ ml/g

6 Hydrodynamic Parameters 2. Partial specific volume: change in volume per gram under specific conditions V g i T, P, g j V i = ( ) Volume = V 1 g 1 + V 2 g 2 Gibbs - Duham relationship We will assume that specific volume is approximately equal to the partial specific volume: V i V i Proteins: V P ~ ml/g Na + DNA: V DNA ~ ml/g

7 3. Effective mass Take into account buoyancy by subtracting the mass of the displaced water [mass - (mass of displaced water)] [m - m(v 2 ρ) = m (1 - V 2 ρ)] M(1 - V 2 ρ) N mv 2 = volume of particle ρ = solvent density V 2 = Vol g of the molecule m = mass of molecule m = (M/N) mv 2 ρ = mass of displaced solvent

8 4. Hydration δ 1 = g H2O g protein m h = m + m δ 1 V h = m h = M ( 1 + δ 1 ) N M ( V 2 + δ 1 V 1 ) N V h = (V 2 + δ 1 V 1 ) Mass of hydrated protein Volume of hydrated protein volume of hydrated protein per gram of protein

9 Problems of hydrodynamic measurements in general-- Properties depend on : 1) size too many parameters! 2) shape 3) solvation Some solutions to obtain useful information: 1 Use more than one property and eliminate one unknown. Example: sedimentation (S o ) and diffusion (D) can eliminate shape factor 2 Work under denaturing conditions to eliminate empirically the shape factor and solvation factors. These are constants for both standards and the unknown sample - then find Molecular weight. Examples: 1) gel filtration of proteins in GuHCl; 2) SDS gels. 3 If Molecular weight is known: compare results with predictions based on an unhydrated sphere of mass M. Then use judgement to explain deviations on the basis of shape or hydration. Behavior of limiting forms - eg; rods, spheres- are calculated for comparisons

10 Next we consider: 1 What effect does the interaction of the hydrodynamic particle and the solvent have on solution properties? viscosity 2 What effect does the particle-solvent interaction have on the motion of the particle? translational motion - free - in a force field rotational motion

11 Viscosity Frictional interactions between layers of solution results in energy dissipation u (velocity) dy A u + du (shearing force) du A du F f = η ( ) A dy dy η coefficient of viscosity cgs units Poise η is related to the amount of energy dissipated per unit volume per unit time

12 Many ways to measure solution viscosity capillary viscometer: measures the rate of flow of solution through a capillary with a pressure drop P

13 Effect of macromolecules on viscosity: only 2 parameters plus solute η η o = 1 + υ φ fraction of volume occupied by particles solvent alone shape factor Shape factor υ axial ratio = a/b For a sphere: υ = 2.5 Shape factor is very large for highly asymmetric molecules b a viscosity effects are very large for very elongated polymers (large ν) (e.g. actin) or molecules that occupy large volume (large φ) (e.g. DNA)

14 Relative viscosity and Specific viscosity 1 η η o = η r relative viscosity = 1 + υφ ν = shape factor η - η o 2 = η sp specific viscosity η o η sp = η r - 1 = υφ φ = fraction of solution volume occupied by solvent particles so But φ = V h c η sp c = υ V h c = g / ml solvent V h = volume per gram of protein occupied by solvated particle ideally, the increase in specific viscosity per gram of solute should be a constant (υ V h ). Often it is not due to non-ideal behavior Shape factor vol/g of protein of hydrated particle

15 Non-ideal case (i.e.,reality): η sp c = υ V h + (const.) c +... η sp c (ideal) c Extrapolate measurement to infinite dilution: Intrinsic viscosity lim (η sp /c) = [η] limit at low concentration (intercept in graph above) C 0 Intrinsic viscosity: [η], units = ml / g [η] = υ V h Intrinsic viscosity is dependent only on size and properties of the isolated macromolecule.

16 Intrinsic viscosity [η] = υ V h monomer ν increases polymer ν = 2.5 [η] = υ V h if ν = 25 then [η] is 10x larger [η] = υ V h if we assume a spherical shape, ν = 2.5, we will calculate an incorrect and very large value for V h

17 For a flexible polymer like DNA or denatured protein Behaves like compact particles with R 0. 8 x radius of gyration (radius of gyration measures mass distribution and can be obtained from light scattering) example: [η] 4 ml / g for a native, globular protein But for a random coil, if there is ~ 100 g bound solvent per gram solute V h V ml/ g [η] = 2.5 V h very large coil

18 Viscosity : examples I Native Hemocyanin ( M = 10 6 ) native vs GuHCl denatured 60 GuHCl denatured η sp C 40 [η] = υ V h = 2.5 V h coiled cm 3 g 20 0 Native C (mg / ml) of BSA compact II T7 DNA (double strand) : salt dependence [η sp ] C cm 3 g expanded coil log [NaCl]

19 Stokes Radius Radius of the sphere which has the hydrodynamic properties consistent With the hydrodynamic measurement 1. measure [η] = υ V h [η] = 2.5 V h 2. assume a sphere 4/3π R s 3 M/N 3. calculate Stokes radius volume of the molecule (spherical) mass of the molecule Solve for the Stokes radius: M 3 R s = [ V h ] N 4 π 1/3 where V h = [η]/2.5

20 MINIMUM RADIUS The radius expected if the molecule is an anhydrous sphere [4/3 π R min3 ] V 2 = (M/N) anhydrous sphere - point of reference M 3 R min = [ N 4π V 2 ] 1/3 CALCULATED

21 Compare the Stokes radius (measured) to the minimum radius expected assuming the molecule is an anhydrous sphere Question: How close does the assumption of an anhydrous sphere come to explaining the value of R s? If R s /R min is not much larger than 1.0, then the assumption of the molecule being spherical is likely reasonable. R s should be slightly larger than R min due to hydration If R s /R min is much larger than 1.0, then either the molecule is not spherical (ν >>2.5) or it is not compact (V h >>V 2 )

22 For an anhydrous particle: [η] = υ V 2 anhydrous vol/gram for an anhydrous sphere: υ = 2.5 [η] = 2.5 V 2 For real macromolecules, the intrinsic viscosity will vary from the above due to 1. correction due to hydration υ = 2.5 V h (hydrated vol/g = V 2 +δ H20 ) 2. correction due to asymmetry υ > 2.5 V h = V 2 (anhydrous vol/g) or both hydration and asymmetry υ > 2.5 and V h = V 2 + δ H2O

23 Comparison of intrinsic viscosity values for two proteins: 1. Ribonuclease: mol wt: 13, 683 V 2 = ml/g R min = 17 Å 2. Collagen: mol wt: 345,000 V 2 = ml/g R min = 59 Å

24 Interpreting Viscosity Data: Does a reasonable amount of hydration explain the measured value of [η]? maximum solvation maximum asymmetry protein [η] ml/g R s (Å) δ H2O (g/g) υ (a/b) Ribonuclease Collagen Stokes Radius [η] = υ V h = υ [4/3 π R s3 ] N M 3 [η] M R s = ( ) 4 π N υ 1/3 2.5 Appropriate for collagen R s /R min = 6.8 (400/59) V h = V 2 shape correction: υ > 2.5 (anhydrous value) hydration correction: V h = V 2 + δ H2O Appropriate for ribonuclease R s /R min = 1.14 (19.3/17) υ = 2.5